Asked by Maho
use logarithmic diff. to find the derivative of the function. Show steps please! so I can see how it is done. Thank you so much!
y=(e^(-x)cos^(2)(x))/(x^(2)+x+1)
y=(e^(-x)cos^(2)(x))/(x^(2)+x+1)
Answers
Answered by
Reiny
y = (e^-x) (cosx)^2 (x^2 + x + 1)
take ln of both sides
ln y = ln e^-x + ln (cosx)^2 + ln(x^2 + x + 1)
= -x + 2 ln(cosx) + ln(x^2 + x + 1)
now differentiate
y' / y = -1 + 2(-sinx/cosx) + (2x+1)/(x^2 + x + 1)
= -1 - 2tanx + (2x+1)/(x^2 + x + 1)
y' = y(-1 - 2tanx + (2x+1)/(x^2 + x + 1))
= [ (e^-x) (cosx)^2 (x^2 + x + 1) ] * [ -1 - 2tanx + (2x+1)/(x^2 + x + 1) ]
sure hope they don't expect us to simplify this
take ln of both sides
ln y = ln e^-x + ln (cosx)^2 + ln(x^2 + x + 1)
= -x + 2 ln(cosx) + ln(x^2 + x + 1)
now differentiate
y' / y = -1 + 2(-sinx/cosx) + (2x+1)/(x^2 + x + 1)
= -1 - 2tanx + (2x+1)/(x^2 + x + 1)
y' = y(-1 - 2tanx + (2x+1)/(x^2 + x + 1))
= [ (e^-x) (cosx)^2 (x^2 + x + 1) ] * [ -1 - 2tanx + (2x+1)/(x^2 + x + 1) ]
sure hope they don't expect us to simplify this
Answered by
Steve
log y = log (e^-x) + log cos^2(x) - log(x^2+x+1)
log y = -x + 2log cos x - log(x^2+x+1)
1/y y' = -1 - 2tanx - (2x+1)/(x^2+x+1)
y' = -(1 + 2tanx + (2x+1)/(x^2+x+1)) * (e^-x cos^2x)/(x^2+x+1)
Now, you can massage that for a few more steps, to get something that pleases you
log y = -x + 2log cos x - log(x^2+x+1)
1/y y' = -1 - 2tanx - (2x+1)/(x^2+x+1)
y' = -(1 + 2tanx + (2x+1)/(x^2+x+1)) * (e^-x cos^2x)/(x^2+x+1)
Now, you can massage that for a few more steps, to get something that pleases you
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.